A tourist drops (from rest) a ping pong ball from the top of the tower, which has a height of 324 meters. Assuming no air resist
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Answer:
The displacement of the air drop after 3 second is 18.27 m.
Explanation:
Mass of the rain drop = m =
Weight of the rain drop = W
Duration of time = t = 3 seconds
Drag force on rain drop =
Motion of the rain drop:
Net force on the rain drop , F= W - D
v = 12.18 m/s
Initial velocity of the rain drop = u = 0 (since, it is starting from rest)
v=u+at (First equation of motion)
(second equation of motion)
s = 18.27 m
The displacement of the air drop after 3 second is 18.27 m.